The vapor pressure of an ideal solution can be predicted using Raoult's law, which states: \[ P_{\text{solution}} = X_X P_X^0 + X_Y P_Y^0 \] Where: - \( P_{\text{solution}} \) is the vapor pressure of the solution, - \( X_X \) and \( X_Y \) are the mole fractions of X and Y, respectively,
- \( P_X^0 \) and \( P_Y^0 \) are the vapor pressures of pure X and Y, respectively.
Given: - Moles of X = 5, Moles of Y = 10, - \( P_X^0 = 63 \, \text{torr}, P_Y^0 = 78 \, \text{torr} \),
- Total moles = 5 + 10 = 15. The mole fractions are: \[ X_X = \frac{5}{15} = \frac{1}{3}, \quad X_Y = \frac{10}{15} = \frac{2}{3} \] Now, applying Raoult's law: \[ P_{\text{solution}} = \left(\frac{1}{3}\right)(63) + \left(\frac{2}{3}\right)(78) \] \[ P_{\text{solution}} = 21 + 52 = 73 \, \text{torr} \] The given vapor pressure is 70 torr, which is lower than the calculated vapor pressure of 73 torr. This means the solution exhibits negative deviation from Raoult's law.
Thus, the solution shows negative deviation.
Therefore, the correct answer is (1) The solution shows negative deviation.
A bob of heavy mass \(m\) is suspended by a light string of length \(l\). The bob is given a horizontal velocity \(v_0\) as shown in figure. If the string gets slack at some point P making an angle \( \theta \) from the horizontal, the ratio of the speed \(v\) of the bob at point P to its initial speed \(v_0\) is :